Java，而使用扫描器时无限循环

Question

我想让用户输入一个String和四个整数值，如果用户输入的类型不匹配，我想让程序继续要求用户输入整数值，为什么代码会一直循环，如果用户插入了错误的类型，永远不会等待用户输入？

import java.util.InputMismatchException;
import java.util.Scanner;

public class Main {
public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    String sessionName;
    int pomoInterv, breakInterv, terminalBreakInterv;
    System.out.println("Please, Enter the session name: ");
    sessionName = scanner.nextLine();
    while (true) {
        try {
            System.out.println("Please, Enter the Pomodoro interval: ");
            pomoInterv = scanner.nextInt();
            break;
        } catch (InputMismatchException ex) {
            System.out.println(ex.getMessage());
        }

    }
    while (true) {
        try {
            System.out.println("Please, Enter the break interval: ");
            breakInterv = scanner.nextInt();
            break;
        } catch (InputMismatchException ex) {
            System.out.println(ex.getMessage());
        }

    }
    while (true){
        try {
            System.out.println("Please, Enter the terminal break interval ");
            terminalBreakInterv = scanner.nextInt();
            break;
        } catch (InputMismatchException ex) {
            System.out.println(ex.getMessage());
        }
    }
    System.out.println("Done!");
    System.out.println(sessionName);
    System.out.println(pomoInterv);
    System.out.println(breakInterv);
    System.out.println(terminalBreakInterv);

}
}

Answer 1

我其实解决了，我只要加一个(scanner . next()；)在下面的catch语句中，是因为Scanner类是如何工作的，参见下面的代码，

import java.util.InputMismatchException;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    String sessionName;
    int pomoInterv, breakInterv, terminalBreakInterv;
    System.out.println("Please, Enter the session name: ");
    sessionName = scanner.nextLine();
    while (true) {
        try {
            System.out.println("Please, Enter the Pomodoro interval: ");
            pomoInterv = scanner.nextInt();
            break;
        } catch (InputMismatchException ex) {
            scanner.next();
            System.out.println(ex.getMessage());
        }

    }
    while (true) {
        try {
            System.out.println("Please, Enter the break interval: ");
            breakInterv = scanner.nextInt();
            break;
        } catch (InputMismatchException ex) {
            scanner.next();
            System.out.println(ex.getMessage());
        }

    }
    while (true){
        try {
            System.out.println("Please, Enter the terminal break interval ");
            terminalBreakInterv = scanner.nextInt();
            break;
        } catch (InputMismatchException ex) {
            scanner.next();
            System.out.println(ex.getMessage());
        }
    }
    System.out.println("Done!");
    System.out.println(sessionName);
    System.out.println(pomoInterv);
    System.out.println(breakInterv);
    System.out.println(terminalBreakInterv);

}
}

检查这个答案https://stackoverflow.com/a/47852703/12565862

Answer 2

当检查用户输入是否正确时，我总是使用do-while循环。

看看以下是否有效：

    Scanner scanner = new Scanner(System.in);
    String sessionName;
    int pomoInterv = -1, breakInterv = -1, terminalBreakInterv = -1;
    System.out.println("Please, Enter the session name: ");
    sessionName = scanner.nextLine();
    boolean isPomoInterv = false;
    do {
        System.out.println("Please, Enter the Pomodoro interval: ");
        String temp = scanner.nextLine();
        try {
            if (Integer.parseInt(temp) > 0) {
                pomoInterv = Integer.parseInt(temp);
                isPomoInterv = true;
            }
        } catch (NumberFormatException e) {
            System.out.println("Try again");
        }

    } while (!isPomoInterv);

    boolean isBreakInterv = false;
    do {
        System.out.println("Please, Enter the break interval: ");
        String temp = scanner.nextLine();
        try {
            if (Integer.parseInt(temp) > 0) {
                breakInterv = Integer.parseInt(temp);
                isBreakInterv = true;
            }
        } catch (NumberFormatException e) {
            System.out.println("Try again");
        }

    } while (!isBreakInterv);

    boolean isTerminalBreakInterv = false;
    do {
        System.out.println("Please, Enter the terminal break interval ");
        String temp = scanner.nextLine();
        try {
            if (Integer.parseInt(temp) > 0) {
                terminalBreakInterv = Integer.parseInt(temp);
                isTerminalBreakInterv = true;
            }
        } catch (NumberFormatException e) {
            System.out.println("Try again");
        }

    } while (!isTerminalBreakInterv);

    System.out.println("Done!");
    System.out.println(sessionName);
    System.out.println(pomoInterv);
    System.out.println(breakInterv);
    System.out.println(terminalBreakInterv);

惊吓袋熊对你的问题的评论是正确的。我用他的评论编写了上面的代码。

Answer 3

根据扫描仪Oracle docs:https://docs . Oracle . com/javase/7/docs/API/Java/util/scanner . html" target="_blank">html

当扫描仪抛出InputMismatchException时，扫描仪不会传递导致异常的令牌，因此可以通过其他方法检索或跳过它。

因此，当您输入整数以外的任何内容时，scanner.nextInt（）不会将其解析为整数并抛出InputMismatchException，并且扫描仪不会移动到下一个标记并一次又一次地尝试读取相同的标记。

要解决这个问题，您可以更改循环，或者使用hasNextInt()方法，或者在catch块中使用scanner.next()，以便扫描器可以移动到下一个标记。